Question

Lead in metal shredder residue. On the basis of data collectedfrom metal shredders across the nation, the amount xof extractable lead in metal shredder residue has an approximateexponential distribution with mean$\mathit{\theta }$= 2.5 milligramsper liter (Florida Shredder’s Association).

a. Find the probability that xis greater than 2 milligramsper liter.

b. Find the probability that xis less than 5 milligrams perliter.

Short answer

a.$P\left(x>2\right)=.0068$

b.$P\left(x<5\right)=0.9999$

Step-by-step solution

Given information

X is an exponential random variable and$\theta =2.5$ .

Define the probability density function

The p.d.f of X is

$$\begin{gathered} f\left(x\right)=\theta e^{-\theta x};x\ge 0 \\ =2.5e^{-2.5x} \end{gathered}$$

Calculate P(x>2)

a

$$\begin{gathered} P\left(x>2\right)=1-P\left(x\le 2\right) \\ =1-{\int }_0^{2}2.5^{-2.5x}dx \\ =1-\left(2.5{\int }_0^2{e}^{-2.5x}dx\right) \\ =1-\left(\frac{5\left({\displaystyle \frac{2}{5}}-{\displaystyle \frac{2e^{-5}}{5}}\right)}{2}\right) \\ =1-\left(e^{-5}\left(e^{-5}-1\right)\right) \\ =1-0.9932 \\ =0.0068 \end{gathered}$$

Hence, the probability of x that x is greater than 2 milligrams per liter is 0.0068.

Calculate P(x<5)

b

$$\begin{gathered} P\left(x<5\right)={\int }_0^{5}2.5e^{-2.5x}dx \\ =\left(2.5{\int }_0^5{e}^{-2.5x}dx\right) \\ =\frac{5\left({\displaystyle \frac{2}{5}}-{\displaystyle \frac{2e^{-{\displaystyle \frac{25}{2}}}}{5}}\right)}{2} \\ =e^{-\frac{25}{2}}\left(e^{-\frac{25}{2}}-1\right) \\ =0.9999 \end{gathered}$$

Hence, the probability of x that x is less than 5 milligrams per liter is 0.9999.